Ira ran out of time while taking a multiple-choice test and plans to guess on the last $6$ questions. Each question has $4$ possible choices, one of which is correct. Let $X=$ the number of answers Ira correctly guesses in the last $6$ questions. What is the probability that he answers fewer than $2$ questions correctly in the last $6$ questions? You may round your answer to the nearest hundredth. $P(X<2)=$
Solution: Strategy (without a fancy calculator) The probability that Ira gets fewer than $2$ questions correct in the $6$ questions is equivalent to the probability that he gets $0$ or $1$ question correct. So we can find those probabilities and add them them together to get our answer: $\begin{aligned} P(X<2)&=P(0\text{ correct})+P(1\text{ correct}) \\\\ &=P(X=0)+P(X=1) \end{aligned}$ Finding $P(X=0)$ There are $4$ possible choices for each question, so we know $P({\text{correct}})={25\%}$ and $P({\text{not}})={75\%}$. Answering $0$ questions correctly is equivalent to answering all $6$ questions incorrectly. We can multiply probabilities since we are assuming independence: $\begin{aligned} P(X=0)&=({0.75})({0.75})({0.75})({0.75})({0.75})({0.75}) \\\\ &=({0.75})^6 \\\\ &\approx0.17798 \end{aligned}$ We'll come back and use this result later. Next, we need to find $P(X=1)$ (the probability that he answers $1$ question correctly). Finding $P(X=1)$ Answering $1$ question correctly in the last $6$ questions means Ira needs to get $1$ question correct and $5$ questions not correct. There are $4$ possible choices for each question, so we know $P({\text{correct}})={25\%}$ and $P({\text{not}})={75\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of getting $1$ question correct followed by $5$ questions not correct: $P({\text{C}}{\text{NNNNN}})=({0.25})({0.75})^5\approx0.05933$ This isn't the entire probability though, because there are other ways to get $1$ question correct from $6$ questions (for example, NNNNNC). How many different ways are there? We can use the combination formula: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _6\text{C}_1&=\dfrac{6!}{(6-1)!\cdot1!} \\\\ &=\dfrac{6 \cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{(\cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}) \cdot 1} \\\\ &=6 \end{aligned}$ There are $6$ ways to get $1$ question correct in $6$ questions. Do they all have the same probability? Each of the $6$ ways has the same probability that we already found: $\begin{aligned} P({\text{C}}{\text{NNNNN}})&=({0.25})({0.75})^5\approx0.05933 \\\\ P({\text{N}}{\text{C}}{\text{NNNN}})&=({0.25})({0.75})^5\approx0.05933 \\\\ P({\text{NN}}{\text{C}}{\text{NNN}})&=({0.25})({0.75})^5\approx0.05933 \\\\ P({\text{NNN}}{\text{C}}{\text{NN}})&=({0.25})({0.75})^5\approx0.05933 \\\\ P({\text{NNNN}}{\text{C}}{\text{N}})&=({0.25})({0.75})^5\approx0.05933 \\\\ P({\text{NNNNN}}{\text{C}})&=({0.25})({0.75})^5\approx0.05933 \end{aligned}$ So we can multiply this probability by $6$ since that is how many ways there are to get $1$ question correct in $6$ questions: $\begin{aligned} P(X=1)&=6(0.25)(0.75)^5 \\\\ &\approx6(0.05933) \\\\ &\approx0.35596 \end{aligned}$ Putting it all together Let's return to our original strategy to answer the question: $\begin{aligned} P(X<2)&=P(0\text{ correct})+P(1\text{ correct}) \\\\ &=P(X=0)+P(X=1) \\\\ &=(0.75)^6+6(0.25)(0.75)^5 \\\\ &\approx0.17798+0.35596 \\\\ &\approx0.53394 \\\\ &\approx0.53 \end{aligned}$ The answer $P(X<2)\approx0.53394\approx0.53$